# Thread: calculus - line integral and work done

1. ## calculus - line integral and work done

question 1 -line integral

Evaluate the integral of C xdy-ydx along the curve C given by the equation y=x^3 from (0,0) to (2,8).

question 2 - work

Find the work done in moving an object through the vector field
F<x,y> = <-y,x> along the upper semicircle x^2 + y^2 =1 from A(1,0) to B(-1,0).

2. Originally Posted by kittycat
question 1 -line integral

Evaluate the integral of C xdy-ydx along the curve C given by the equation y=x^3 from (0,0) to (2,8).
you can parametrize.. let x=t so y=t^3, then the limits of t will be from 0 to 2.. so, dx=dt and dy=3t^2 dt. therefore
$\displaystyle xdy-ydx = 3t^3 dt - t^3 dt = 2t^3 dt \,$ so,
$\displaystyle \int_0^2 {2t^3 \, dt} = \frac{1}{2}t^4$ evaluated from 0 to 2 equals 8..

Originally Posted by kittycat
question 2 - work

Find the work done in moving an object through the vector field
F<x,y> = <-y,x> along the upper semicircle x^2 + y^2 =1 from A(1,0) to B(-1,0).
so you have F(x,y) = -yi + xj

let x=t, $\displaystyle y= \sqrt{1-t^2}, \, -1 \leq t \leq 1$

so,

$\displaystyle Work = \int_C Fdr = \int_C {-ydx + xdy} = \int_{-1}^1 {-\sqrt{1-t^2} dt + t \frac{-t}{\sqrt{1-t^2}}dt}$

$\displaystyle = \int_{-1}^1 \frac {t^2-1 -t^2}{\sqrt{1-t^2}}dt = \int_{-1}^1 \frac {-dt}{\sqrt{1-t^2}}$

i think, you can do the rest.. Ü

3. Originally Posted by kalagota
you can parametrize.. let x=t so y=t^3, then the limits of t will be from 0 to 2.. so, dx=dt and dy=3t^2 dt. therefore
$\displaystyle xdy-ydx = 3t^3 dt - t^3 dt = 2t^3 dt \,$ so,
$\displaystyle \int_0^2 {2t^3 \, dt} = \frac{1}{2}t^4$ evaluated from 0 to 2 equals 8..

so you have F(x,y) = -yi + xj

let x=t, $\displaystyle y= \sqrt{1-t^2}, \, -1 \leq t \leq 1$

so,

$\displaystyle Work = \int_C Fdr = \int_C {-ydx + xdy} = \int_{-1}^1 {-\sqrt{1-t^2} dt + t \frac{-t}{\sqrt{1-t^2}}dt}$

$\displaystyle = \int_{-1}^1 \frac {t^2-1 -t^2}{\sqrt{1-t^2}}dt = \int_{-1}^1 \frac {-dt}{\sqrt{1-t^2}}$

i think, you can do the rest.. Ü
This is correct, but a detail has been omitted. The definition of work is
$\displaystyle W = \int_C F \cdot ds$
not
$\displaystyle Work = \int_C Fdr$
(The missing dot product is incorrect, and the dr implies a radial differential.)

So
$\displaystyle W = \int_C F \cdot ds = \int_C (-y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = \int_C -ydx + xdy$
as kalagota stated.

-Dan

4. well, i just omitted the dot.. anyways, according to my notes (a year ago Ü), the definition states

$\displaystyle W= \int_C F \cdot Tds$

where $\displaystyle T=\frac {dr}{ds}\, \,$ the tangential vector, that is why i came with

$\displaystyle \int_C F \cdot dr$..

anyways, thanks!!

5. Originally Posted by kalagota
well, i just omitted the dot.. anyways, according to my notes (a year ago Ü), the definition states

$\displaystyle W= \int_C F \cdot Tds$

where $\displaystyle T=\frac {dr}{ds}\, \,$ the tangential vector, that is why i came with

$\displaystyle \int_C F \cdot dr$..

anyways, thanks!!
I do it that way because I'm a Physicist, not a Mathematician.

Either way, the point is you get to the correct answer.

-Dan

6. Originally Posted by topsquark
I do it that way because I'm a Physicist, not a Mathematician.

Either way, the point is you get to the correct answer.

-Dan
hehe.. thanks!!

7. well, i just omitted the dot.. anyways, according to my notes (a year ago Ü), the definition states

where the tangential vector, that is why i came with

..

=================================

Hi Kalagota,

My lecture note states that T is a unit tangent vector where T(t) = r'(t)/IIr'(t)II .

Does this mean the same as what you mentioned --- the tangential vector?

8. Originally Posted by kittycat
well, i just omitted the dot.. anyways, according to my notes (a year ago Ü), the definition states

where the tangential vector, that is why i came with

..

=================================

Hi Kalagota,

My lecture note states that T is a unit tangent vector where T(t) = r'(t)/IIr'(t)II .

Does this mean the same as what you mentioned --- the tangential vector?
yes, they are just the same. the reason is that, you gave the function in terms of the parameter t; while mine was parameter s with respect to the arc length..
actually,
$\displaystyle ||r'(t)|| = s'(t)$
then from here, you will that they are the same.. Ü

9. Hi Kalagota,

Could you please tell me how to solve the following question? Thank you very much.

question (work )
Evaluate the integral c of F dot dr where F(x,y,z) = 8x^2yz i+ 5zj-4xyk and c is the curve given by r(t)=ti + t^2j+t^3k where 0 < or = t< or = 1.

10. Originally Posted by kittycat
Hi Kalagota,

Could you please tell me how to solve the following question? Thank you very much.

question (work )
Evaluate the integral c of F dot dr where F(x,y,z) = 8x^2yz i+ 5zj-4xyk and c is the curve given by r(t)=ti + t^2j+t^3k where 0 < or = t< or = 1.
recall, if $\displaystyle \bold{F}$ is a continuous vector field on a smooth curve $\displaystyle C$ given by the vector function $\displaystyle \bold{r} (t)$, $\displaystyle a \le t \le b$. Then the line integral of $\displaystyle \bold{F}$ along $\displaystyle C$ is given by:

$\displaystyle \int_C \bold{F} \cdot d\bold{r} = \int_a^b \bold{F}( \bold{r}(t)) \cdot \bold{r}'(t)~dt = \int_C \bold{F} \cdot \bold{T}~ds$

use the middle one